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A house has well-insulated walls. It contains a volume of \(100 \mathrm{m}^{3}\) of air at \(300 \mathrm{K}\). (a) Calculate the energy required to increase the temperature of this diatomic ideal gas by \(1.00^{\circ} \mathrm{C} .\) (b) What If? If this energy could be used to lift an object of mass \(m\) through a height of \(2.00 \mathrm{m},\) what is the value of \(m ?\)

Short Answer

Expert verified
The energy required to increase the temperature of the ideal diatomic gas by \(1.00^{\circ} \mathrm{C}\) is \(1.20 \cdot 10^{6} \, Joules\), and the mass that could be lifted by this energy is approximately \(61225 \, kg\).

Step by step solution

01

Calculate the energy required to increase the temperature

To calculate the energy required to increase the temperature of the ideal diatomic gas by \(1.00^{\circ} \mathrm{C}\), we can use the formula for the change in internal energy which is \( \Delta U = nC_v\Delta T = \frac{pV}{R}\Delta T \). Given that \(V = 100 \, \mathrm{m}^{3}\), \(T = 300 \, \mathrm{K}\), \(\Delta T = 1.00 \, \mathrm{K}\) (as 1 degree Celsius change is equivalent to 1 Kelvin change), \(R = 8.31 \, J/(mol \cdot K)\), and \(p \) is atmospheric pressure (\(1.00 \cdot 10^{5} \, P\)), we can substitute these values into the formula to find \(\Delta U = \frac{(1.00 \cdot 10^{5} \cdot 100)}{8.31} \cdot 1.00 = 1.20 \cdot 10^{6} \, J\). So, the energy required is \(1.20 \cdot 10^{6} \, Joules\).
02

Calculate the value of mass, \(m\)

Now in part (b) of the question, the problem states that this energy can be used to lift an object of mass \(m\) through a height of \(2.00 \, m\). This is a basic work-energy problem, with work being defined as \(W = mgh\), where \(g\) is the acceleration due to gravity, \(h\) is the height and \(m\) is the mass of the object. Setting the work equal to the energy obtained in part (a), we can solve for \(m\) as \(m = \frac{W}{gh} = \frac{1.20 \cdot 10^{6}}{9.8 \cdot 2.00} = 61225 \, kg\). Therefore, the value of \(m\) that could be lifted by this energy is approximately \(61225 \, kg\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diatomic Ideal Gas
When talking about diatomic ideal gases, it’s important to understand that these are gases composed of molecules that consist of two atoms. Common examples include oxygen (O₂) and nitrogen (N₂), which make up a substantial part of the Earth's atmosphere. In the realm of thermodynamics, diatomic gases are pivotal because they exhibit more complex behaviors than monatomic gases due to their rotational and vibrational energy states.

In terms of thermodynamic calculations, like in the given exercise dealing with temperature change, diatomic ideal gases typically have a higher specific heat capacity compared to their monatomic counterparts. This is because they require more energy to achieve the same temperature change, due to their additional degrees of freedom. These degrees of freedom refer to the ways in which the molecules can store energy, namely through translational, rotational, and at high temperatures, vibrational modes.

The understanding of these properties is crucial when performing calculations related to energy changes in systems involving diatomic gases.
Internal Energy
Internal energy is a concept intrinsic to understanding heat and thermodynamics. It refers to the total energy contained within a system, encompassing both kinetic energy from particle motion and potential energy from interactions between particles. In the context of gases, internal energy is largely due to the kinetic energy of atom and molecule movements.

A key point to remember is that for ideal gases, internal energy is a function primarily of temperature. This is because pressure and volume changes do not alter the energy of individual molecules, except by changing the temperature. The formula used in the exercise, \( \Delta U = nC_v\Delta T \), directly ties changes in internal energy \( \Delta U \) to temperature changes \( \Delta T \), with \( C_v \) being the molar specific heat at constant volume.

This relationship simplifies the prediction of energy changes in diatomic gases, as seen in our exercise where we calculated the energy needed to heat a certain volume of air.
Work-Energy Principle
The work-energy principle is a foundational concept in physics, stating that the work done on an object is equal to the change in its kinetic energy. This principle is applicable in many areas, including mechanics and thermodynamics.

In step 2 of the original solution, this principle helps calculate the mass of an object that can be lifted using a specific amount of energy. Here, the work performed by the energy is calculated using the equation \( W = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height through which the object is moved.

Understanding this relationship allows one to translate thermal energy transformations into mechanical work scenarios. This is helpful in diverse scientific fields, from engineering applications to environmental systems.
Specific Heat Capacity
Specific heat capacity is a valuable property that defines the relationship between the heat added to a substance and the resultant temperature change. Mathematically, it is represented as \( C = \frac{Q}{m\Delta T} \), where \( Q \) is the heat added, \( m \) is the mass, and \( \Delta T \) is the temperature change.

For diatomic ideal gases, specific heat capacity at constant volume \( C_v \) or constant pressure \( C_p \) is higher than for monatomic gases. This is because the additional molecular behavior (like rotation) in diatomic molecules requires greater energy for a change in temperature. In our exercise, this concept played a role in calculating the energy needed to increase the temperature of an air volume by 1°C.

This property aids in predicting how substances respond to thermodynamic processes and is key to understanding energy balance in systems.

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Most popular questions from this chapter

Use the definition of Avogadro's number to find the mass of a helium atom.

Consider 2.00 mol of an ideal diatomic gas. (a) Find the total heat capacity of the gas at constant volume and at constant pressure assuming the molecules rotate but do not vibrate. (b) What If? Repeat, assuming the molecules both rotate and vibrate.

Show that the mean free path for the molecules of an ideal gas is $$\ell=\frac{k_{\mathrm{B}} T}{\sqrt{2} \pi d^{2} P}$$ where \(d\) is the molecular diameter.

Calculate the change in internal energy of 3.00 mol of helium gas when its temperature is increased by \(2.00 \mathrm{K}.\)

(a) Show that the speed of sound in an ideal gas is $$v=\sqrt{\frac{\gamma R T}{M}}$$ where \(M\) is the molar mass. Use the general expression for the speed of sound in a fluid from Section \(17.1,\) the definition of the bulk modulus from Section \(12.4,\) and the result of Problem 59 in this chapter. As a sound wave passes through a gas, the compressions are either so rapid or so far apart that thermal conduction is prevented by a negligible time interval or by effective thickness of insulation. The compressions and rarefactions are adiabatic. (b) Compute the theoretical speed of sound in air at \(20^{\circ} \mathrm{C}\) and compare it with the value in Table \(17.1 .\) Take \(M=\) \(28.9 \mathrm{g} / \mathrm{mol} .\) (c) Show that the speed of sound in an ideal gas is $$v=\sqrt{\frac{\gamma k_{\mathrm{B}} T}{m}}$$ where \(m\) is the mass of one molecule. Compare it with the most probable, average, and rms molecular speeds.

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